# Special values of the dilgoarithm function

Prove that

$$\mathrm{Li}_2\left( \frac{\sqrt{5}-1}{2} \right) = \frac{\pi^2}{10} – \log^2 \left( \frac{\sqrt{5}-1}{2}\right)$$

$$\textit{proof}$$

Use the following functional equation

$$\mathrm{Li}_2 \left( \frac{z}{z-1} \right) + \frac{1}{2} \mathrm{Li}_2 (z^2) – \mathrm{Li}_2(-z) = -\frac{1}{2} \log^2 (1-z)$$

These are proved here and here
Now let $z = \frac{1-\sqrt{5}}{2}$

$$z^2 = \frac{1-2\sqrt{5}+5}{4}= \frac{3-\sqrt{5}}{2} \, \implies \, \frac{z}{z-1} = \frac{\sqrt{5}-1}{1+\sqrt{5}} = \frac{3-\sqrt{5}}{2}$$

Hence we have

$$\frac{3}{2} \mathrm{Li}_2 \left( \frac{3-\sqrt{5}}{2} \right) – \mathrm{Li}_2 \left( \frac{\sqrt{5}-1}{2} \right) = -\frac{1}{2} \log^2 \left( \frac{\sqrt{5}+1}{2}\right)$$

We already established the following functional equation

$$\mathrm{Li}_2 (z) + \mathrm{Li}_2 (1-z) = \frac{\pi^2}{6} – \log(z) \log(1-z)$$

Put  $z = \frac{3-\sqrt{5}}{2}$

$$\mathrm{Li}_2 \left( \frac{3-\sqrt{5}}{2}\right) + \mathrm{Li}_2 \left(\frac{\sqrt{5}-1}{2}\right) = \frac{\pi^2}{6} – \log\left(\frac{3-\sqrt{5}}{2} \right) \log \left(\frac{\sqrt{5}-1}{2}\right)$$

Solving the two representations for  $\mathrm{Li}_2 \left(\frac{\sqrt{5}-1}{2}\right)$ we get our result.

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